A body is projected vertically upwards with a initial velocity of 30m/s. it rises with a deceleration of 10m/s. find it’s height above the point of projection of anytime t. 

The height of a body projected vertically upwards can be determined using the kinematic equation:

\[
h(t) = v_0 t - \frac{1}{2} a t^2
\]

Where:
- \( h(t) \) is the height above the point of projection at time \( t \),
- \( v_0 \) is the initial velocity (30 m/s in this case),
- \( a \) is the acceleration due to gravity (deceleration is \( a = 10 \, \text{m/s}^2 \)),
- \( t \) is the time in seconds.

Substituting the values into the equation:

\[
h(t) = 30t - \frac{1}{2}(10)t^2
\]

Simplifying:

\[
h(t) = 30t - 5t^2
\]

Thus, the height of the body at any time \( t \) is given by:

\[
h(t) = 30t - 5t^2 \, \text{meters}
\]

This equation represents the height of the body above the point of projection as a function of time.

Do you have any questions or want further details?

### Related Questions:
1. How would you find the time at which the body reaches its maximum height?
2. What is the maximum height the body will reach?
3. How long will it take for the body to return to the point of projection?
4. What is the velocity of the body at any time \( t \)?
5. What would happen if the initial velocity were doubled?

### Tip:
Remember that the velocity becomes zero at the maximum height of a vertically projected body.

A body is projected vertically upwards with an initial velocity of 30 m/s. It rises with a deceleration of 10 m/s². Find its height above the point of projection at any time t.

Discover how to find the height of a body projected vertically upwards with an initial velocity of 30 m/s, rising with a deceleration of 10 m/s². This solution explores kinematic equations and projectile motion, suitable for students in Grades 9-12.

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